The equation of the ellipse whose vertices ar | vedclass

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Question

(a) Vertices $( \pm 5,\,0) \equiv ( \pm a,\,0)$ $&rArr; a = 5$

Foci $( \pm 4,\,0) \equiv ( \pm ae,\,0)$

$e = \frac{4}{5}$, 

$b = (5)\,\l

Vedclass · Accepted Answer

$9{x^2} + 25{y^2} = 225$

Vedclass · Answer

(a) Vertices $( \pm 5,\,0) \equiv ( \pm a,\,0)$ $&rArr; a = 5$

Foci $( \pm 4,\,0) \equiv ( \pm ae,\,0)$

$e = \frac{4}{5}$, 

$b = (5)\,\left( {\frac{3}{5}} \right) = 3$

Hence equation is $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$

$i.e.$, $9{x^2} + 25{y^2} = 225$.

The equation of the ellipse whose vertices are $( \pm 5,\;0)$ and foci are $( \pm 4,\;0)$ is

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